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Question: Let Z be a non real complex no. lying on the circle \|Z\| = 1, then Z is :...

Let Z be a non real complex no. lying on the circle |Z| = 1, then Z is :

A

tan(argZ2)\left( \frac{\arg Z}{2} \right)

B

sin(argZ2)\left( \frac{\arg Z}{2} \right)

C

1+itan(argZ2)1itan(argZ2)\frac{1 + i\tan\left( \frac{\arg Z}{2} \right)}{1 - i\tan\left( \frac{\arg Z}{2} \right)}

D

None of these

Answer

1+itan(argZ2)1itan(argZ2)\frac{1 + i\tan\left( \frac{\arg Z}{2} \right)}{1 - i\tan\left( \frac{\arg Z}{2} \right)}

Explanation

Solution

Sol. Since |Z| = 1

let Z = cosa + iisina

(where a is argument of Z)

Ž Z = 1tan2(α/2)1+tan2(α/2)\frac{1 - \tan^{2}(\alpha/2)}{1 + \tan^{2}(\alpha/2)}+ ii 2tan(α/2)1+tan2(α/2)\frac{2\tan(\alpha/2)}{1 + \tan^{2}(\alpha/2)}

Ž Z = 1tan2(α/2)+2itan(α/2)1+tan2(α/2)\frac{1 - \tan^{2}(\alpha/2) + 2i\tan(\alpha/2)}{1 + \tan^{2}(\alpha/2)}

Ž Z = (1+itanα/2)2(1itanα/2)(1+itanα/2)\frac{\left( 1 + i\tan\alpha/2 \right)^{2}}{(1 - i\tan\alpha/2)(1 + i\tan\alpha/2)}

Ž Z =

Ž (1+itan(α/2)1itan(α/2))\left( \frac{1 + i\tan(\alpha/2)}{1 - i\tan(\alpha/2)} \right)

Ž Z = 1+itan(Arg. Z2)1itan(Arg. Z2) \frac{1 + itan\left( \frac{Arg.\ Z}{2} \right)}{1–itan\left( \frac{Arg.\ Z}{2} \right)\ }